Integrand size = 15, antiderivative size = 99 \[ \int \sqrt [4]{b x^3+a x^4} \, dx=\frac {(b+4 a x) \sqrt [4]{b x^3+a x^4}}{8 a}+\frac {3 b^2 \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{16 a^{7/4}}-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{16 a^{7/4}} \]
1/8*(4*a*x+b)*(a*x^4+b*x^3)^(1/4)/a+3/16*b^2*arctan(a^(1/4)*x/(a*x^4+b*x^3 )^(1/4))/a^(7/4)-3/16*b^2*arctanh(a^(1/4)*x/(a*x^4+b*x^3)^(1/4))/a^(7/4)
Time = 0.38 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.17 \[ \int \sqrt [4]{b x^3+a x^4} \, dx=\frac {x^{9/4} (b+a x)^{3/4} \left (2 a^{3/4} x^{3/4} \sqrt [4]{b+a x} (b+4 a x)+3 b^2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )-3 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )\right )}{16 a^{7/4} \left (x^3 (b+a x)\right )^{3/4}} \]
(x^(9/4)*(b + a*x)^(3/4)*(2*a^(3/4)*x^(3/4)*(b + a*x)^(1/4)*(b + 4*a*x) + 3*b^2*ArcTan[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)] - 3*b^2*ArcTanh[(a^(1/4)*x ^(1/4))/(b + a*x)^(1/4)]))/(16*a^(7/4)*(x^3*(b + a*x))^(3/4))
Time = 0.31 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.43, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {1910, 1930, 1938, 73, 854, 827, 216, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sqrt [4]{a x^4+b x^3} \, dx\) |
| \(\Big \downarrow \) 1910 |
| \(\displaystyle \frac {1}{8} b \int \frac {x^3}{\left (a x^4+b x^3\right )^{3/4}}dx+\frac {1}{2} x \sqrt [4]{a x^4+b x^3}\) |
| \(\Big \downarrow \) 1930 |
| \(\displaystyle \frac {1}{8} b \left (\frac {\sqrt [4]{a x^4+b x^3}}{a}-\frac {3 b \int \frac {x^2}{\left (a x^4+b x^3\right )^{3/4}}dx}{4 a}\right )+\frac {1}{2} x \sqrt [4]{a x^4+b x^3}\) |
| \(\Big \downarrow \) 1938 |
| \(\displaystyle \frac {1}{8} b \left (\frac {\sqrt [4]{a x^4+b x^3}}{a}-\frac {3 b x^{9/4} (a x+b)^{3/4} \int \frac {1}{\sqrt [4]{x} (b+a x)^{3/4}}dx}{4 a \left (a x^4+b x^3\right )^{3/4}}\right )+\frac {1}{2} x \sqrt [4]{a x^4+b x^3}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{8} b \left (\frac {\sqrt [4]{a x^4+b x^3}}{a}-\frac {3 b x^{9/4} (a x+b)^{3/4} \int \frac {\sqrt {x}}{(b+a x)^{3/4}}d\sqrt [4]{x}}{a \left (a x^4+b x^3\right )^{3/4}}\right )+\frac {1}{2} x \sqrt [4]{a x^4+b x^3}\) |
| \(\Big \downarrow \) 854 |
| \(\displaystyle \frac {1}{8} b \left (\frac {\sqrt [4]{a x^4+b x^3}}{a}-\frac {3 b x^{9/4} (a x+b)^{3/4} \int \frac {\sqrt {x}}{1-a x}d\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}}{a \left (a x^4+b x^3\right )^{3/4}}\right )+\frac {1}{2} x \sqrt [4]{a x^4+b x^3}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle \frac {1}{8} b \left (\frac {\sqrt [4]{a x^4+b x^3}}{a}-\frac {3 b x^{9/4} (a x+b)^{3/4} \left (\frac {\int \frac {1}{1-\sqrt {a} \sqrt {x}}d\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}}{2 \sqrt {a}}-\frac {\int \frac {1}{\sqrt {a} \sqrt {x}+1}d\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}}{2 \sqrt {a}}\right )}{a \left (a x^4+b x^3\right )^{3/4}}\right )+\frac {1}{2} x \sqrt [4]{a x^4+b x^3}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{8} b \left (\frac {\sqrt [4]{a x^4+b x^3}}{a}-\frac {3 b x^{9/4} (a x+b)^{3/4} \left (\frac {\int \frac {1}{1-\sqrt {a} \sqrt {x}}d\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}\right )}{a \left (a x^4+b x^3\right )^{3/4}}\right )+\frac {1}{2} x \sqrt [4]{a x^4+b x^3}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{8} b \left (\frac {\sqrt [4]{a x^4+b x^3}}{a}-\frac {3 b x^{9/4} (a x+b)^{3/4} \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}\right )}{a \left (a x^4+b x^3\right )^{3/4}}\right )+\frac {1}{2} x \sqrt [4]{a x^4+b x^3}\) |
(x*(b*x^3 + a*x^4)^(1/4))/2 + (b*((b*x^3 + a*x^4)^(1/4)/a - (3*b*x^(9/4)*( b + a*x)^(3/4)*(-1/2*ArcTan[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)]/a^(3/4) + A rcTanh[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)]/(2*a^(3/4))))/(a*(b*x^3 + a*x^4) ^(3/4))))/8
3.14.79.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 1)/n) Subst[Int[x^m/(1 - b*x^n)^(p + (m + 1)/n + 1), x], x, x/(a + b*x^n )^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, - 2^(-1)] && IntegersQ[m, p + (m + 1)/n]
Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[x*((a*x^j + b*x^n)^p/(n*p + 1)), x] + Simp[a*(n - j)*(p/(n*p + 1)) Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && !IntegerQ[p] && LtQ[0, j, n] && GtQ[p, 0] && NeQ[n*p + 1, 0]
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1))) I nt[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && Gt Q[m + j*p - n + j + 1, 0] && NeQ[m + n*p + 1, 0]
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^IntPart[m]*(c*x)^FracPart[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(F racPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])) Int[x^(m + j* p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !Inte gerQ[p] && NeQ[n, j] && PosQ[n - j]
Time = 2.39 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.14
| method | result | size |
| pseudoelliptic | \(\frac {16 a^{\frac {7}{4}} \left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}} x +4 b \,a^{\frac {3}{4}} \left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}-3 \ln \left (\frac {-a^{\frac {1}{4}} x -\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}\right ) b^{2}-6 \arctan \left (\frac {\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) b^{2}}{32 a^{\frac {7}{4}}}\) | \(113\) |
1/32*(16*a^(7/4)*(x^3*(a*x+b))^(1/4)*x+4*b*a^(3/4)*(x^3*(a*x+b))^(1/4)-3*l n((-a^(1/4)*x-(x^3*(a*x+b))^(1/4))/(a^(1/4)*x-(x^3*(a*x+b))^(1/4)))*b^2-6* arctan(1/a^(1/4)/x*(x^3*(a*x+b))^(1/4))*b^2)/a^(7/4)
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.34 \[ \int \sqrt [4]{b x^3+a x^4} \, dx=-\frac {3 \, a \left (\frac {b^{8}}{a^{7}}\right )^{\frac {1}{4}} \log \left (\frac {3 \, {\left (a^{2} \left (\frac {b^{8}}{a^{7}}\right )^{\frac {1}{4}} x + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{2}\right )}}{x}\right ) - 3 \, a \left (\frac {b^{8}}{a^{7}}\right )^{\frac {1}{4}} \log \left (-\frac {3 \, {\left (a^{2} \left (\frac {b^{8}}{a^{7}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{2}\right )}}{x}\right ) - 3 i \, a \left (\frac {b^{8}}{a^{7}}\right )^{\frac {1}{4}} \log \left (-\frac {3 \, {\left (i \, a^{2} \left (\frac {b^{8}}{a^{7}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{2}\right )}}{x}\right ) + 3 i \, a \left (\frac {b^{8}}{a^{7}}\right )^{\frac {1}{4}} \log \left (-\frac {3 \, {\left (-i \, a^{2} \left (\frac {b^{8}}{a^{7}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{2}\right )}}{x}\right ) - 4 \, {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} {\left (4 \, a x + b\right )}}{32 \, a} \]
-1/32*(3*a*(b^8/a^7)^(1/4)*log(3*(a^2*(b^8/a^7)^(1/4)*x + (a*x^4 + b*x^3)^ (1/4)*b^2)/x) - 3*a*(b^8/a^7)^(1/4)*log(-3*(a^2*(b^8/a^7)^(1/4)*x - (a*x^4 + b*x^3)^(1/4)*b^2)/x) - 3*I*a*(b^8/a^7)^(1/4)*log(-3*(I*a^2*(b^8/a^7)^(1 /4)*x - (a*x^4 + b*x^3)^(1/4)*b^2)/x) + 3*I*a*(b^8/a^7)^(1/4)*log(-3*(-I*a ^2*(b^8/a^7)^(1/4)*x - (a*x^4 + b*x^3)^(1/4)*b^2)/x) - 4*(a*x^4 + b*x^3)^( 1/4)*(4*a*x + b))/a
\[ \int \sqrt [4]{b x^3+a x^4} \, dx=\int \sqrt [4]{a x^{4} + b x^{3}}\, dx \]
\[ \int \sqrt [4]{b x^3+a x^4} \, dx=\int { {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (79) = 158\).
Time = 0.28 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.45 \[ \int \sqrt [4]{b x^3+a x^4} \, dx=\frac {\frac {6 \, \sqrt {2} b^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}} a} + \frac {6 \, \sqrt {2} b^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}} a} + \frac {3 \, \sqrt {2} b^{3} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x}}\right )}{\left (-a\right )^{\frac {3}{4}} a} + \frac {3 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{3} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x}}\right )}{a^{2}} + \frac {8 \, {\left ({\left (a + \frac {b}{x}\right )}^{\frac {5}{4}} b^{3} + 3 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} a b^{3}\right )} x^{2}}{a b^{2}}}{64 \, b} \]
1/64*(6*sqrt(2)*b^3*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x)^( 1/4))/(-a)^(1/4))/((-a)^(3/4)*a) + 6*sqrt(2)*b^3*arctan(-1/2*sqrt(2)*(sqrt (2)*(-a)^(1/4) - 2*(a + b/x)^(1/4))/(-a)^(1/4))/((-a)^(3/4)*a) + 3*sqrt(2) *b^3*log(sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x))/(( -a)^(3/4)*a) + 3*sqrt(2)*(-a)^(1/4)*b^3*log(-sqrt(2)*(-a)^(1/4)*(a + b/x)^ (1/4) + sqrt(-a) + sqrt(a + b/x))/a^2 + 8*((a + b/x)^(5/4)*b^3 + 3*(a + b/ x)^(1/4)*a*b^3)*x^2/(a*b^2))/b
Time = 6.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.38 \[ \int \sqrt [4]{b x^3+a x^4} \, dx=\frac {4\,x\,{\left (a\,x^4+b\,x^3\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {7}{4};\ \frac {11}{4};\ -\frac {a\,x}{b}\right )}{7\,{\left (\frac {a\,x}{b}+1\right )}^{1/4}} \]